Question

\[\int\frac{5 x^2 + 20x + 6}{x^3 + 2 x^2 + x} dx\]

Answer 1

We have,

\[I = \int\frac{5 x^2 + 20x + 6}{x^3 + 2 x^2 + x}\]

\[ = \int\frac{\left( 5 x^2 + 20x + 6 \right) dx}{x \left( x^2 + 2x + 1 \right)}\]

\[ = \int\frac{\left( 5 x^2 + 20x + 6 \right) dx}{x \left( x + 1 \right)^2}\]

\[\text{Let }\frac{5 x^2 + 20x + 6}{x \left( x + 1 \right)^2} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{\left( x + 1 \right)^2}\]

\[ \Rightarrow \frac{5 x^2 + 20x + 6}{x \left( x + 1 \right)^2} = \frac{A \left( x + 1 \right)^2 + B \left( x \right) \left( x + 1 \right) + C \left( x \right)}{x \left( x + 1 \right)^2}\]

\[ \Rightarrow 5 x^2 + 20x + 6 = A \left( x^2 + 2x + 1 \right) + B \left( x^2 + x \right) + Cx\]

\[ \Rightarrow 5 x^2 + 20x + 6 = \left( A + B \right) x^2 + \left( 2A + B + C \right) x + A\]

\[\text{Equating coefficients of like terms}\]

\[A + B = 5 . . . . . \left( 1 \right)\]

\[2A + B + C = 20 . . . . . \left( 2 \right)\]

\[ A = 6 . . . . . \left( 3 \right)\]

\[\text{Solving (1), (2) and (3), we get}\]

\[A = 6 \])

\[B = - 1\]

\[C = 9\]

\[ \therefore \frac{5 x^2 + 20x + 6}{x \left( x + 1 \right)^2} = \frac{6}{x} - \frac{1}{x + 1} + \frac{9}{\left( x + 1 \right)^2}\]

\[ \Rightarrow I = 6\int\frac{dx}{x} - \int\frac{dx}{x + 1} + 9\int\frac{dx}{\left( x + 1 \right)^2}\]

\[ = 6 \log \left| x \right| - \log \left| x + 1 \right| - \frac{9}{x + 1} + C\]